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Summary Engineering Mechanics
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A summary of all lectures of engineering mechanics (statics) of the first period including the weekly test completely worked out.
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Trabajo colaborativo 3 de Estática, ejercicios "PROBLEMAS" del libro de Ingeniería Mecánica Estática, R.C. Hibbeler, edición 14
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Solution Manual - Engineering Mechanics Statics 12th Edition By RCHibbeler Chapter 2
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AE1130-IEngineering Mechanics
AE1130-I
Delft University of Technology
1
, AE1130-I
Table of Contents
Forces and Vectors……………………………………………………………………………………….…………………………………….3
Force Concept………………………………………………………………………………………………………………………………….…5
Varignon’s Principle……………………………………………………………………………………………………..…………………….9
Example Forces and Vectors, Force Concept and Varignon’s Principle……………………………………………..12
Mechanical Systems…………………………………………………………………………………………………………………..…….13
Forces and Moments in 3D……………………………………………………………………………………………………………….18
Example Mechanical Systems and Forces and Moments in 3D………………………………………………………….22
Forces and Moments in 3D……………………………………………………………………………………………………………….23
Trusses……………………………………………………………………………………………………………………………………………..28
Example Forces and Moments in 3D and Trusses……………………………………………………………………………..33
Frames…………………………………………………………………………………………………………………..…………………………34
Distributed Loads……………………………………………………………………………………………….…………………………….39
Centre of Mass………………………………………………………………………………………………………………………………….42
Example Frames, Distributed Loads and Centre of Mass……………………………………………………….………….46
Moment of Inertia………………………………………………………………………………………………………………….………..48
Virtual Work……………………………………………………………………………………………………………………………………..52
Example Moment of Inertia and Virtual Work…………………………………………………………………………………..60
Internal Effects in Beams……………………………………………………………………………………………………………..…..61
N, V, M-Diagrams……………………………………………………………………………………………………………………………..68
Example Internal Effects in Beams and N, V, M-Diagrams………………………………………………………..……….72
Internal Force Diagrams and Frames………………………………………………………………………………………………..74
Example Internal Force Diagrams and Frames…………………………………….………………………………………..….81
2
, AE1130-I
Forces and Vectors
Mechanics
Mechanics deals with models for the behavior of bodies under the action of forces.
Force
Every force has:
- A magnitude
- A direction
- A point of application
Vector
- Mathematical quantity
- Has magnitude & direction
Three classes of vectors
- Free vector (only has a magnitude and a direction, but no point of application)
- Sliding vector (has a magnitude and a direction, but a point of application which slides over a
line)
- Fixed vector (has a fixed magnitude, fixed direction and a fixed point of application)
Conventions ⃗
𝑉
- V in textbook
- ⃗ or 𝑉
In writing 𝑉 ⃗ or 𝑉̅ or 𝑉
θ
Mathematical rules
- ⃗ its opposite is −𝑉
1st rule: 𝑉 ⃗
nd
- 2 rule: vectors can be added using the parallelogram rule ⃗⃗⃗
𝑉1
⃗ = ⃗⃗⃗
𝑉 𝑉1 + ⃗⃗⃗
𝑉2
|𝑉| ≠ |𝑉1 | + |𝑉2 |
- ⃗ = ⃗⃗⃗
3rd rule: 𝑉 𝑉1 − ⃗⃗⃗
𝑉2
⃗⃗⃗
𝑉2
Set of axes ⃗
𝑉
𝑦 𝑉𝑦
𝜃 = arctan ( )
𝑉𝑥
⃗
𝑉
⃗⃗⃗
𝑉𝑦 ⃗ = 𝑉𝑛⃗
𝑉 |𝑛| = 1
θ 𝑛 = cos 𝜃 𝑖 + 𝑠𝑖𝑛𝜃𝑗
𝑥
⃗⃗⃗𝑥
𝑉 1 0 0
𝑖 = (0) 𝑗 = (1) 𝑘⃗ = (0)
0 0 1
3
, AE1130-I
Set of axes extra
𝑦 𝑉𝑥 = 𝑉 cos 𝜃
⃗ 𝑉𝑦 = 𝑉 sin 𝜃
𝑉
𝑉𝑦 𝑗
𝑉 = |𝑉| = √𝑉𝑥2 + 𝑉𝑦2
𝑗 θ 𝑥
𝑖 𝑉𝑥 𝑖
Example
⃗
𝐹 = 𝐴+ 𝐵
12 kN
𝐴 = |𝐴|𝑛⃗
𝐴
𝐴 = 𝐴𝑥 𝑖 + 𝐴𝑦 𝑗
θ = 30 ̊
𝐴 = |𝐴| cos 𝜃 𝑖 + |𝐴|𝑠𝑖𝑛𝜃𝑗
4
⃗ = 𝐵𝑛⃗
𝐵
3 ⃗
𝐵
⃗ = 𝐵𝑥 𝑖 + 𝐵𝑦 𝑗
𝐵
20 kN
3 4
⃗ = |𝐵|𝑖 − |𝐵|𝑗
𝐵
5 5
⃗
𝐹 = 𝐴+ 𝐵
3 4
𝐹 = (|𝐴| cos 𝜃 + 5 |𝐵|) 𝑖 + (|𝐴|𝑠𝑖𝑛𝜃 − 5 |𝐵|)𝑗
1 3 1 4
𝐹 = (12 ∙ √3 + ∙ 20) 𝑖 + (12 ∙ − ∙ 20)𝑗
2 5 2 5
𝐹 = (6√3 + 12)𝑖 − 10𝑗
2
|𝐹| = √(6√3 + 12) + (−10)2 = 24.52 𝑘𝑁
Scalar product (dot product)
𝐴 ∙ 𝐵⃗ = 𝑠𝑐𝑎𝑙𝑎𝑟 𝑎𝑛𝑠𝑤𝑒𝑟
𝐴 ∙ 𝐵⃗ = 𝐴𝑥 𝐵𝑥 + 𝐴𝑦 𝐵𝑦 + 𝐴𝑧 𝐵𝑧
𝐴 ∙ 𝐵⃗ = 𝐵⃗ ∙ 𝐴
𝑖 ∙ 𝑖 = 𝑗 ∙ 𝑗 = 𝑘⃗ ∙ 𝑘⃗ = 1
𝑖 ∙ 𝑗 = 𝑗 ∙ 𝑖 = 𝑗 ∙ 𝑘⃗ = 𝑘⃗ ∙ 𝑗 = 𝑘⃗ ∙ 𝑖 = 𝑖 ∙ 𝑘⃗ = 0
𝑃⃗ ∙ 𝑄⃗ = (𝑃𝑥 𝑖 + 𝑃𝑦 𝑗 + 𝑃𝑧 𝑘⃗ ) ∙ (𝑄𝑥 𝑖 + 𝑄𝑦 𝑗 + 𝑄𝑧 𝑘⃗ )
𝑃⃗ ∙ 𝑄⃗ = 𝑃𝑥 𝑄𝑥 + 𝑃𝑦 𝑄𝑦 + 𝑃𝑧 𝑄𝑧
𝑃⃗ ∙ 𝑃⃗ = 𝑃𝑥2 + 𝑃𝑦2 + 𝑃𝑧2
𝑃⃗ ∙ 𝑄⃗
cos 𝜃 =
|𝑃| ∙ |𝑄|
4
, AE1130-I
Force Concept
1
Forces example 3
Decompose the force shown into the axes provided.
𝐹
1
1 eb
𝑒𝑎 =
⃗⃗⃗⃗ (−1𝑖 + 1𝑗) 1 3
√2 ea
1 1 5
𝑒𝑎 = − √2𝑖 + √2𝑗
⃗⃗⃗⃗
2 2
1
𝑒𝑏 =
⃗⃗⃗⃗ (1𝑖 + 3𝑗)
√10
1 3
𝑒𝑏 =
⃗⃗⃗⃗ √10𝑖 + √10𝑗
10 10
𝐹 = 5𝑖 + 3𝑗
𝐹 = 𝐹𝑎 ⃗⃗⃗⃗
𝑒𝑎 + 𝐹𝑏 ⃗⃗⃗⃗
𝑒𝑏
1 1 1 3
𝐹 = 𝐹𝑎 (− √2𝑖 + √2𝑗) + 𝐹𝑏 ( √10𝑖 + √10𝑗)
2 2 10 10
1 1 1 3
𝐹 = (− √2𝐹𝑎 + √10𝐹𝑏 ) 𝑖 + ( √2𝐹𝑎 + √10𝐹𝑏 )𝑗
2 10 2 10
1 1
x-direction → − 2 √2𝐹𝑎 + 10 √10𝐹𝑏 = 5
1 3
y-direction → 2 √2𝐹𝑎 + 10 √10𝐹𝑏 = 3
4
Addition: 10 √10𝐹𝑏 = 8
𝐹𝑏 = 2√10 𝑁
Fill in 𝐹𝑏 → 𝐹𝑎 = −3√2 𝑁
Sailplane
𝐹
𝐹
⃗⃗⃗ = 𝑚𝑔
𝑊
⃗⃗⃗ = 𝑚𝑔
𝑊
Diagram Free-body-diagram
5
, AE1130-I
Sailplane example
𝑦′
𝑦 𝐹
𝐹
α θ 𝑥
𝑥′
⃗⃗⃗ = 𝑚𝑔
𝑊
⃗⃗⃗ = 𝑚𝑔
𝑊
Diagram Free-body-diagram
Given: α = 10 ,̊ θ = 80 ,̊ m = 300 kg, |𝐹| = 3000 𝑁 and g = 9.81 m/s2.
𝑅⃗ = 𝐹 + 𝑊⃗⃗⃗
𝐹 cos 𝜃 𝑚𝑔 sin 𝛼
𝑅⃗ = ( )+( )
𝐹𝑠𝑖𝑛𝜃 −𝑚𝑔 cos 𝛼
3000 cos 80 300 ∙ 9.81 ∙ 𝑔 sin 10 1.03 ∙ 103
𝑅⃗ = ( )+( )=( )𝑁
3000𝑠𝑖𝑛80 −300 ∙ 9.81 ∙ cos 10 56.1
Newton’s laws
- 1st law: a particle remains at rest or continues to move with uniform velocity (in a straight
line with a constant speed) if there is no unbalanced force acting on it.
0
𝑅⃗ = 0⃗ = ( ) → ∑ 𝐹𝑥 = 0 𝑎𝑛𝑑 ∑ 𝐹𝑦 = 0
0
- 2nd law: the acceleration of a particle is proportional to the vector sum of forces acting on it,
and it is in the direction of this vector sum.
∑ 𝐹 = 𝑚𝑎
⃗ → ∑𝐹 = 0
𝑎=0
Approach to a problem
- Formulate the problem
- state data
- state desired result
- state assumptions and approximations
- Developing a solution
- draw diagram (free-body-diagram)
- state governing principles
- make calculations
- check on accuracy
- check for consistent units
- check reasonability
- draw conclusion
6
, AE1130-I
Airplane example
𝐹
⃗
𝑇 θ
⃗⃗⃗
𝑊
W = 12000 N
θ = 65 ̊
What is F and what is T?
𝐹
𝑦
⃗
𝑇 θ 𝑥
⃗⃗⃗
𝑊
∑𝐹 = 0
∑ 𝐹𝑦↑+ : 0 = 𝐹 sin 𝜃 − 𝑚𝑔
𝐹 sin 𝜃 = 𝑚𝑔
𝑚𝑔 12000
𝐹= = = 13240.5 𝑁
sin 𝜃 sin 65
∑ 𝐹𝑥→+ : 0 = −𝑇 + 𝐹 cos 𝜃
𝑚𝑔 cos 𝜃 𝑚𝑔 12000
𝑇 = 𝐹 cos 𝜃 = = = = 5595.6 𝑁
sin 𝜃 tan 𝜃 tan 65
Plausibility is OK.
7
, AE1130-I
Example blimp
L
Wi
F
wind
θ
L = 2000 N, m = 150 kg, Wi = 100 N (windforce). What is 𝜃 and what is F?
𝑦
L
𝑥
Wi θ
F
𝑊𝑖
∑ 𝐹𝑥→+ : 0 = −𝑊𝑖 + 𝐹 cos 𝜃 → 𝐹 =
cos 𝜃
∑ 𝐹𝑦↑+ : 0 = 𝐿 − 𝑚𝑔 − 𝐹 sin 𝜃
𝑊𝑖
𝐿 = 𝑚𝑔 + ∙ sin 𝜃 = 𝑚𝑔 + 𝑊𝑖 tan 𝜃
cos 𝜃
𝐿 − 𝑚𝑔 2000 − 1471.5
tan 𝜃 = = = 5.285
𝑊𝑖 100
𝜃 = 79.285 ̊
𝑊𝑖 100
𝐹= = = 537.88 𝑁
cos 𝜃 0.186
Moments
LT
L d
𝑀 = 𝐿𝑇 𝑑
8
, AE1130-I
Varignon’s Principle
Varignon’s principle
The moment of a force about any point is equal to the sum of the moments of the components of
this force.
𝑀 = 𝑑𝐹
Right-hand rule
If you want to know whether a moment is positive or negative, use your right hand and put your
fingers in the direction of the arm. After that bend your fingers in the direction of the force. If your
thumb points out of the paper the moment is positive but if it points into the paper the moment is
negative.
Example
Calculate
the
moments
in point 1,
2, 3 and 4.
3 2
4
1
Point 1: 𝑀1 = 16 ∙ 100 = 1600 𝑘𝑁𝑚
Point 2: 𝑀2 = −16 ∙ 100 = −1600 𝑘𝑁𝑚
Point 3: 𝑀3 = −16 ∙ 100 = −1600 𝑘𝑁𝑚
Point 4: 𝑀4 = 0 ∙ 100 = 0 𝑘𝑁𝑚
Example
𝐹 = 𝐹 cos 𝛼 𝑖 + 𝐹 sin 𝛼 𝑗
𝑀 = 𝐹𝑥 𝑑𝑦 + 𝐹𝑦 𝑑𝑥
𝑑 = 𝑙 cos 𝛽 𝑖 + 𝑙 sin 𝛽 𝑗
𝑀 = −𝐹 cos 𝛼 𝑙 sin 𝛽 + 𝐹 sin 𝛼 𝑙 cos 𝛽
9
, AE1130-I
Example
𝑀𝐴 = −𝐹𝑎 + 𝐹(𝑎 + 𝑑) = 𝐹𝑑
Cross product
𝑟 𝐹
⃗⃗ = 𝑟 × 𝐹 = ( 𝑥 ) × ( 𝑥 ) = 𝑟𝑥 𝐹𝑦 − 𝑟𝑦 𝐹𝑥
𝑀
𝑟𝑦 𝐹𝑦
𝑟×𝐹 ≠𝐹×𝑟
𝑟 × 𝐹 = −𝐹 × 𝑟
𝑢1 𝑣1
⃗ = (𝑢2 ) ; 𝑣 = (𝑣2 )
𝑢
𝑢3 𝑣3
⃗ × 𝑣 = (𝑢2 𝑣3 − 𝑢3 𝑣2 )𝑖 + (𝑢3 𝑣1 − 𝑢1 𝑣3 )𝑗 + (𝑢1 𝑣2 − 𝑢2 𝑣1 )𝑘⃗
𝑢
Resultant force
A single resultant force in a specific point that replaces all forces and couples in a body.
𝑅 = ∑𝐹
𝑀𝑅 = ∑(𝑟 × 𝐹) = 0
Example
𝑦
𝐿⃗ ⃗⃗⃗⃗
𝐿𝑇
𝑥
2 ⃗
𝐷
⃗
𝑇
⃗⃗⃗
𝑊
2 23
L = 2.4 MN, LT = 0.3 MN, W = 2.5 MN, T = 0.8 MN, D = 0.7 MN. What is 𝑅⃗ and what is 𝑇
⃗?
𝑛
𝑅⃗ = ∑ 𝐹
⃗⃗𝑖
𝑖=1
𝑛
𝑅𝑥 = ∑ ⃗⃗⃗⃗
𝐹𝑥𝑖 = 𝐷 − 𝑇 = 0.7 − 0.8 = −0.1 𝑀𝑁
𝑖=1
𝑛
𝑅𝑦 = ∑ ⃗⃗⃗⃗⃗
𝐹𝑦𝑖 = 𝐿 + 𝐿 𝑇 − 𝑊 = 2.4 + 0.3 − 2.5 = 0.2 𝑀𝑁
𝑖=1
−0.1
𝑅⃗ = ( ) 𝑀𝑁
0.2
𝑀0↺+ = −2𝑇 + 0 + 0 − 2𝑊 + 25𝐿 𝑇 = −2 ∙ 0.8 − 2 ∙ 2.5 + 25 ∙ 0.3 = 0.9 𝑀𝑁𝑚
10
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